Hydraulics / fluid dynamics

[fayan77]

Homework Statement

Homework Equations

y₁+P₁/$\gamma$+$\alpha$v₁²/2g = y₂+P₂/$\gamma$+$\alpha$v₂²/2g + h_L

The Attempt at a Solution

so it gives me pressure at p2 but not p1. I tried using Bernoulli's equation to find p1 i did plugged into energy equation above and got an answer but it did not make sense i got h_L = .01 ft (I assumed that v1 = v2 but i do not think that is valid)

 

[haruspex]

got an answer

Please post your working (and not as an image - those are for textbook extracts and diagrams).

 

[fayan77]

...That is the problem exercise. Do you want me to type it instead?
I would like some direction so I can try it and post my work. Am I correct to use Bernoulli's equation to find P₁? and will velocities be the same in both sections?

 

[haruspex]

That is the problem exercise. Do you want me to type it instead?

The image you posted is fine. I am asking you to post your own working, but not as an image. I find that if I do not say that up front the working gets posted as an image which is hard to read and even harder to comment on.

Am I correct to use Bernoulli's equation to find P1? and will velocities be the same in both sections?

Yes and yes.

 

[fayan77]

ok, So I start off with Bernoulli's equation solving for P₁. Because v₁ = v₂ the kinetic energy term will cancel out leaving us with

P₁ + $\rho$gh₁ = P₂ + $\rho$gh₂
P₁ +(1.94)(32.2)(80) = 2592 + (1.94)(32.3)(12)
P₁ = -1655.8 psf
This does not make sense to me ( having a negative pressure at the top of the system)
Anyways now I will use the energy equation assume $\alpha$ = 1 also because v₁ = v₂ velocity head will cancel out from equation
y₁ + P₁/$\gamma$ +$\alpha$v₁²/2g = y₂ + P₂/$\gamma$ +$\alpha$v₂²/2g + h_L
80 + (-1655.8/62.4) = 12 + (2592/62.4) + h_L
h_L = -.07 ft
This does not make sense to me (negative head) also I am certain that I am approaching this the wrong way because I did not use all the givens in the problem like Q, and L of pipe, and diameter of the pipe.

 

[haruspex]

I start off with Bernoulli's equation

Ok, I misunderstood what you meant by that in post #1.
Bernoulli's equation is an energy equation, but it assumes no losses. I took your "relevant equation" to be a modified Bernoulli that allowed for losses. However, I am not familiar with this form, so please explain what α and γ represent, or post a link.
I would take the inlet pressure to be atmospheric and the outlet pressure to be 18 psi higher. I am not sure what the height info means, but it sounds like there is a drop of 68 ft along the pipe. That's a bit over two atmospheres, so with no flow one would expect a pressure difference of over 30 psi.
It is not clear what form of answer is required... is it the percentage loss or rate of loss?

 

[fayan77]

Divide Bernoulli's equation by $\gamma$ and include head losses. Everything is in terms of head (length) $\gamma$ = $\rho$g alpha is coefficient correction

 

[haruspex]

Divide Bernoulli's equation by $\gamma$ and include head losses. Everything is in terms of head (length) $\gamma$ = $\rho$g alpha is coefficient correction

Ok, that's about what I guessed. Does the "coefficient correction" represent the the loss due to drag? Or does h_L represent energy loss?